Question 14 Marks
If $x=(4-\sqrt{15})$, find the values of $x+\frac{1}{x}$
Answer$x+\frac{1}{x}$
$ x+\frac{1}{x}=(4-\sqrt{15})+\frac{1}{(4-\sqrt{15})}$
$=\frac{(4-\sqrt{15})^2+1}{(4-\sqrt{15})} $
$ =\frac{16+15-8 \sqrt{15}+1}{(4-\sqrt{15})} $
$ =\frac{8(4-\sqrt{15})}{(4-\sqrt{15})} $
$ =8$
View full question & answer→Question 24 Marks
If $x=(4-\sqrt{15})$, find the values of $\frac{1}{x}$
Answer$\frac{1}{x}$
$ \frac{1}{x}=\frac{1}{(4-\sqrt{15})}$
$ =\frac{1}{(4-\sqrt{15})} \times \frac{(4+\sqrt{15})}{(4+\sqrt{15})} $
$ =\frac{(4+\sqrt{15})}{16-15} $
$=(4+\sqrt{15})$
View full question & answer→Question 34 Marks
Simplify the following$\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$
Answer$\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$
$=\frac{(\sqrt{5}+\sqrt{3})^2+(\sqrt{5}-\sqrt{3})^2}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}$
$=\frac{5+3+\sqrt{15}+5+3-\sqrt{15}}{5-3} $
$=\frac{16}{2}$
$ =8$
View full question & answer→Question 44 Marks
Simplify the following$\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}-\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}$
Answer$\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}-\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}} $
$ =\frac{(\sqrt{7}-\sqrt{3})^2-(\sqrt{7}+\sqrt{3})^2}{(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})} $
$ =\frac{7+3-2 \sqrt{21}-7-3-2 \sqrt{21}}{(\sqrt{7})^2-(\sqrt{3})^2} $
$ =\frac{-4 \sqrt{21}}{7-3} $
$ =-\sqrt{21}$
View full question & answer→Question 54 Marks
Simplify the following$\frac{\sqrt{5}-2}{\sqrt{5}+2}-\frac{\sqrt{5}+2}{\sqrt{5}-2}$
Answer$\frac{\sqrt{5}-2}{\sqrt{5}+2}-\frac{\sqrt{5}+2}{\sqrt{5}-2}$
$ =\frac{(\sqrt{5}-2)^2-(\sqrt{5}+2)^2}{(\sqrt{5}+2)(\sqrt{5}-2)}$
$ =\frac{5+4-4 \sqrt{5}-5-4-4 \sqrt{5}}{(\sqrt{5})^2-(2)^2} $
$=\frac{-8 \sqrt{5}}{5-4}$
$ =-8 \sqrt{5}$
View full question & answer→Question 64 Marks
Simplify the following$\frac{3}{5-\sqrt{3}}+\frac{2}{5+\sqrt{3}}$
Answer$\frac{3}{5-\sqrt{3}}+\frac{2}{5+\sqrt{3}}$
$=\frac{3(5+\sqrt{3})+2(5-\sqrt{3})}{(5-\sqrt{3})(5+\sqrt{3})}$
$=\frac{15+3 \sqrt{3}+10-2 \sqrt{3}}{(5)^2-(\sqrt{3})^2}$
$=\frac{25+\sqrt{3}}{25-3}$
$=\frac{25+\sqrt{3}}{22}$
View full question & answer→Question 74 Marks
Simplify by rationalising the denominator in the following.$\frac{7 \sqrt{3}-5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$
Answer$ \frac{7 \sqrt{3}-5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$
$= \frac{7 \sqrt{3}-5 \sqrt{2}}{\sqrt{48}+\sqrt{18}} \times \frac{\sqrt{48}-\sqrt{18}}{\sqrt{48}-\sqrt{18}}$
$= \frac{7 \sqrt{144}-7 \sqrt{54}-5 \sqrt{96}+5 \sqrt{36}}{(\sqrt{48})^2-(\sqrt{18})^2}$
$= \frac{84-21 \sqrt{6}-20 \sqrt{6}+30}{48-18}$
$= \frac{144-41 \sqrt{6}}{30}$
View full question & answer→Question 84 Marks
Simplify by rationalising the denominator in the following.$\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}}$
Answer$\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}}$
$=\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}} \times \frac{3 \sqrt{5}+2 \sqrt{6}}{3 \sqrt{5}+2 \sqrt{6}}$
$=\frac{6 \sqrt{30}+24-15-2 \sqrt{30}}{(3 \sqrt{5})^2-(2 \sqrt{6})^2}$
$=\frac{6 \sqrt{30}+9-2 \sqrt{30}}{45-24}$
$=\frac{4 \sqrt{30}+9}{21}$
View full question & answer→Question 94 Marks
Simplify by rationalising the denominator in the following.$\frac{4+\sqrt{8}}{4-\sqrt{8}}$
Answer$\frac{4+\sqrt{8}}{4-\sqrt{8}} $
$=\frac{4+\sqrt{8}}{4-\sqrt{8}} \times \frac{4+\sqrt{8}}{4+\sqrt{8}} $
$ =\frac{(4+\sqrt{8})^2}{(4)^2-(\sqrt{8})^2}$
$ =\frac{16+8+8 \sqrt{8}}{16-18}$
$ =\frac{24+8 \sqrt{8}}{8} $
$=3+\sqrt{8}$
View full question & answer→Question 104 Marks
Simplify by rationalising the denominator in the following.$\frac{5+\sqrt{6}}{5-\sqrt{6}}$
Answer$\frac{5+\sqrt{6}}{5-\sqrt{6}} $
$ =\frac{5+\sqrt{6}}{5-\sqrt{6}} \times \frac{5+\sqrt{6}}{5+\sqrt{6}} $
$=\frac{(5+\sqrt{6})^2}{(5)^2-(\sqrt{6})^2}$
$=\frac{25+6+10 \sqrt{6}}{25-6} $
$=\frac{31+10 \sqrt{6}}{19}$
View full question & answer→Question 114 Marks
Simplify by rationalising the denominator in the following.$\frac{3-\sqrt{3}}{2+\sqrt{2}}$
Answer$\frac{3-\sqrt{3}}{2+\sqrt{2}} $
$=\frac{3-\sqrt{3}}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}}$
$ =\frac{3(2-\sqrt{2})-\sqrt{3}(2-\sqrt{2})}{(2)^2-(\sqrt{2})^2} $
$ =\frac{6-3 \sqrt{2}-2 \sqrt{3}+\sqrt{6}}{4-2} $
$=\frac{6-3 \sqrt{2}-2 \sqrt{3}+\sqrt{6}}{2}$
View full question & answer→Question 124 Marks
Simplify by rationalising the denominator in the following.$\frac{2}{3+\sqrt{7}}$
Answer$\frac{2}{3+\sqrt{7}} $
$=\frac{2}{3+\sqrt{7}} \times \frac{3-\sqrt{7}}{3-\sqrt{7}} $
$=\frac{2(3-\sqrt{7})}{(3)^2-(\sqrt{7})^2} $
$ =\frac{2(3-\sqrt{7})}{9-7} $
$=\frac{2(3-\sqrt{7})}{2}$
$ =3-\sqrt{7}$
View full question & answer→Question 134 Marks
Simplify by rationalising the denominator in the following.$\frac{1}{\sqrt{3}+\sqrt{2}}$
Answer$\frac{1}{\sqrt{3}+\sqrt{2}} $
$ =\frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} $
$ =\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2} $
$=\frac{\sqrt{3}-\sqrt{2}}{3-2}$
$ =\frac{\sqrt{3}-\sqrt{2}}{1} $
$ =\sqrt{3}-\sqrt{2}$
View full question & answer→Question 144 Marks
Simplify by rationalising the denominator in the following.$\frac{1}{5+\sqrt{2}}$
Answer$\frac{1}{5+\sqrt{2}} $
$ =\frac{1}{5+\sqrt{2}} \times \frac{5-\sqrt{2}}{5-\sqrt{2}}$
$ =\frac{5-\sqrt{2}}{(5)^2-(\sqrt{2})^2} $
$ =\frac{5-\sqrt{2}}{25-2} $
$ =\frac{5-\sqrt{2}}{23}$
View full question & answer→Question 154 Marks
Write the following in descending order:$5 \sqrt{3}, \sqrt{15}$ and $3 \sqrt{5}$
AnswerSince $5 \sqrt{3}=\sqrt{5^2 \times 3}=\sqrt{25 \times 3}=\sqrt{75}$
$\sqrt{15}=\sqrt{15} $
$3 \sqrt{5}=\sqrt{3^2 \times 5}=\sqrt{9 \times 5}=\sqrt{45}$
Since, $75>45>15$,
we have $\sqrt{75}>\sqrt{45}>\sqrt{15}$.
Hence, $5 \sqrt{3}>3 \sqrt{5}>\sqrt{15}$.
View full question & answer→Question 164 Marks
Write the following in descending order:$\sqrt{2}, \sqrt[3]{5}$ and $\sqrt[4]{10}$
AnswerSince $\sqrt{2}=2^{\frac{1}{2}}$ has power $\frac{1}{2}$,
$\sqrt[3]{5}=5^{\frac{1}{3}}$ has power $\frac{1}{3}$
$\sqrt[4]{10}=10^{\frac{1}{4}}$ has power $\frac{1}{4}$
Now,$ \text{L.C.M.}$ of $2,3$ and $4=12$
$\therefore \sqrt{2}=2^{\frac{1}{2}}=2^{\frac{6}{12}}=\left(2^6\right)^{\frac{1}{12}}=(64)^{\frac{1}{12}}$
$\sqrt[3]{5}=5^{\frac{1}{2}}=5^{\frac{4}{12}}=\left(5^4\right)^{\frac{1}{12}}=(625)^{\frac{1}{12}}$
$\sqrt[4]{10}=10^{\frac{1}{4}}=10^{\frac{3}{12}}=\left(10^3\right)^{\frac{1}{12}}=(1000)^{\frac{1}{12}}$
Since, $1000>625>64$,
we have $(1000)^{\frac{1}{12}}>(625)^{\frac{1}{12}}>(64)^{\frac{1}{12}}$.
Hence, $\sqrt[4]{10}>\sqrt[3]{5}>\sqrt{2}$.
View full question & answer→Question 174 Marks
Write the following in ascending order:$7 \sqrt[3]{5}, 6 \sqrt[3]{4}$ and $5 \sqrt[3]{6}$
AnswerSince $7 \sqrt[3]{5}=\sqrt[3]{7^3 \times 5}=\sqrt[3]{343 \times 5}=\sqrt[3]{1715} $
$6 \sqrt[3]{4}=\sqrt[3]{6^3 \times 4}=\sqrt[3]{216 \times 4}=\sqrt[3]{864} $
$5 \sqrt[3]{6}=\sqrt[3]{5^3 \times 4}=\sqrt[3]{125 \times 6}=\sqrt[3]{750}$
Since, $750<864<1715$,
we have $\sqrt[3]{750}<\sqrt[3]{864}<\sqrt[3]{1715}$.
Hence, $5 \sqrt[3]{6}<6 \sqrt[3]{4}<7 \sqrt[3]{5}$.
View full question & answer→Question 184 Marks
Write the following in ascending order:$5 \sqrt{7}, 7 \sqrt{5}$ and $6 \sqrt{2}$
Answer$5 \sqrt{7}=\sqrt{5^2 \times 7}=\sqrt{25 \times 7}=\sqrt{175} $
$7 \sqrt{5}=\sqrt{7^2 \times 5}=\sqrt{49 \times 5}=\sqrt{245} $
$6 \sqrt{2}=\sqrt{6^2 \times 2}=\sqrt{36 \times 2}=\sqrt{72}$
Since, $72<175<245$,
we have $\sqrt{72}<\sqrt{175}<\sqrt{245}$.
Hence, $6 \sqrt{2}<5 \sqrt{7}<7 \sqrt{5}$.
View full question & answer→Question 194 Marks
Write the following in ascending order:$2 \sqrt[3]{3}, 4 \sqrt[3]{3}$ and $3 \sqrt[3]{3}$
AnswerSince $2 \sqrt[3]{3}=\sqrt[3]{2^3 \times 3}=\sqrt[3]{8 \times 3}=\sqrt[3]{24} $
$4 \sqrt[3]{3}=\sqrt[3]{4^3 \times 3}=\sqrt[3]{64 \times 3}=\sqrt[3]{192}$
$3 \sqrt[3]{3}=\sqrt[3]{3^3 \times 3}=\sqrt[3]{27 \times 3}=\sqrt[3]{81}$
Since, $24<81<192$,
we have $\sqrt[3]{24}<\sqrt[3]{81}<\sqrt[3]{192}$.
Hence, $2 \sqrt[3]{3}<3 \sqrt[3]{3}<4 \sqrt[3]{3}$.
View full question & answer→Question 204 Marks
Compare the following:$\sqrt[3]{48}$ and $\sqrt{36}$
Answer$\sqrt[3]{48}=48^{\frac{1}{3}}$ has power $\frac{1}{3} $
$\sqrt{36}=6$
Now,$ \text{L.C.M.}$ of $3$ and $1=3$
$\sqrt[3]{48}=48^{\frac{1}{3}} $
$\sqrt{36}=6=6^{\frac{3}{3}}=\left(6^3\right)^{\frac{1}{3}}=216^{\frac{1}{3}}$
Since $48<216$,
we have $48^{\frac{1}{3}}<216^{\frac{1}{3}}$
Hence, $\sqrt[3]{48}<\sqrt{36}$.
View full question & answer→Question 214 Marks
Compare the following:$\sqrt[4]{12}$ and $\sqrt[3]{15}$
Answer$\sqrt[4]{12}=12^{\frac{1}{4}}$ has power $\frac{1}{4} $
$\sqrt[3]{15}=15^{\frac{1}{3}}$ has power $\frac{1}{3}$
Now, $\text{L.C.M}$. of $4$ and $3=12$
$\sqrt[4]{12}=12^{\frac{1}{4}}=12^{\frac{3}{12}}=\left(12^3\right)^{\frac{1}{12}}=(1728)^{\frac{1}{12}} $
$\sqrt[3]{15}=15^{\frac{1}{3}}=15^{\frac{4}{12}}=\left(15^4\right)^{\frac{1}{12}}=(50625)^{\frac{1}{12}}$
Since $1728<50625$,
we have $(1728)^{\frac{1}{12}}<(50625)^{\frac{1}{12}}$.
Hence, $\sqrt[4]{12}<\sqrt[3]{15}$.
View full question & answer→Question 224 Marks
Write four rational numbers between $\sqrt{2}$ and $\sqrt{3}$
AnswerSince squares of $\sqrt{2}$ and $\sqrt{3}$ are $2$ and $3$ respectively.
Now, find four rational numbers between $2$ and $3$ such that each of them is a perfect square.
Let the numbers be $2.25,2.4025,2.56,2.89$,
where,
$\sqrt{2.25}=1.5$
$\sqrt{2.4025}=1.55$
$\sqrt{2.56}=1.6$
$\sqrt{2.89}=1.7$
Hence, required rational numbers between $\sqrt{2}$ and $\sqrt{3}$ are $1.5,1.55,1.6$ and $1.7 .$
View full question & answer→Question 234 Marks
Write two rational numbers between $\sqrt{3}$ and $\sqrt{7}$
AnswerSince square of $\sqrt{3}$ and $\sqrt{7}$ are $3$ and $7$ respectively.
Now, find two rational numbers between $3$ and $7$ such that each of them is a perfect square.
Let the number be $4$ and $5.76 ,$
where,
$\sqrt{4}=2$$\sqrt{5.76}=2.4$
Hence, required rational numbers between $\sqrt{3}$ and $\sqrt{7}$ are $2$ and $2.4 .$
View full question & answer→Question 244 Marks
Find the greatest and the smallest rational number among the following.$\frac{6}{7}, \frac{9}{14}$ and $\frac{23}{28}$
AnswerGiven number : $\frac{6}{7}, \frac{9}{14}$ and $\frac{23}{28}$
The $\text{L.C.M.}$ of $7,14$ and $28$ is $28.$
Thus, number are :
$ \frac{6}{7}=\frac{6 \times 4}{7 \times 4}$
$=\frac{24}{28} ; \frac{9}{14}$
$=\frac{9 \times 2}{14 \times 2}$
$=\frac{18}{28}$ and $\frac{23}{28} . $
Since $24>23>18$,
we have $\frac{6}{7}>\frac{23}{28}>\frac{9}{14}$.
Hence, the greatest rational number is $\frac{6}{7}$ and
the smallest rational number is $\frac{9}{14}$.
View full question & answer→Question 254 Marks
Express the following decimal as a rational number.$0.89$
AnswerLet $x=0 . \overline{89}$
Then, $x=0.898989 \ldots$
Here, the number of digits recurring is $2,$
so we multiply both sides of the equation $(1)$ by $100 .$
$\therefore 100 x=100 \times 0.898989 \ldots $
$=89.8989 \ldots \ldots . \ldots(2)$
On subtracting $(1)$ from $(2),$ we get
$99 x=89 $
$\therefore x=\frac{89}{99} $
$\therefore 0 . \overline{89}$
$=\frac{89}{99}$
View full question & answer→Question 264 Marks
Arrange the following rational numbers in descending order.$\frac{-3}{8}, \frac{2}{5}$ and $\frac{-1}{3}$
AnswerGiven numbers: $\frac{-3}{8}, \frac{2}{5}$ and $\frac{-1}{3}$
The $\text{L.C.M.}$ of $8,5$ and $3$ is $120 .$
Thus, numbers are :
$\frac{-3}{8} $
$=\frac{-3 \times 15}{8 \times 15} $
$=\frac{-45}{120} \text {. } $
$\frac{2}{5} $
$=\frac{2 \times 24}{5 \times 24} $
$=\frac{48}{120} \text {. } $
$\frac{-1}{3} $
$=\frac{-1 \times 40}{3 \times 40} $
$=\frac{-40}{120} $
Since $48>-40>-45$,
we have $\frac{2}{5}>\frac{-1}{3}>\frac{-3}{8}$.
View full question & answer→Question 274 Marks
Arrange the following rational numbers in descending order.$\frac{-7}{10}, \frac{-8}{15}$ and $\frac{-11}{30}$
AnswerGiven numbers: $\frac{-7}{10}, \frac{-8}{15}$ and $\frac{-11}{30}$
The $\text{L.C.M.} $of $10,15$ and $30$ is $30.$
Thus, numbers are :
$\frac{-7}{10} $
$=\frac{-7 \times 3}{10 \times 3} $
$=\frac{-21}{30} $
$\frac{-8}{15} $
$=\frac{-8 \times 2}{15 \times 2} $
$=\frac{-6}{30} $
$\frac{-11}{30}$
Since $-11>-16>-21$,
we have $\frac{-11}{30}>\frac{-8}{15}>\frac{-7}{10}$.
View full question & answer→Question 284 Marks
Arrange the following rational numbers in descending order.$\frac{4}{3}, \frac{-14}{5}$ and $\frac{17}{15}$
AnswerGiven numbers: $\frac{4}{3}, \frac{-14}{5}$ and $\frac{17}{15}$
The $\text{L.C.M.} $of $3$ and $5$ is $15.$
Thus, numbers are :
$\frac{4}{3}$
$=\frac{4 \times 3}{3 \times 5}$
$=\frac{20}{15}$
$\frac{-14}{5}$
$=\frac{-14 \times 3}{5 \times 3}$
$=\frac{-42}{15}$
$\frac{17}{15}$
Since $20>17>-42$,
we have $\frac{4}{3}>\frac{17}{5}>\frac{-14}{5}$.
View full question & answer→